If + F (x) is a even function (- F, x) = 0, then (f) (x) is equal to (- 0, f) = 0=

If + F (x) is a even function (- F, x) = 0, then (f) (x) is equal to (- 0, f) = 0=

From F (x + 2) = - f (x), f (2012) = - f (2010), f (2010) = - f (2008), so f (2012) = f (2008) = f (2004) =. = f (0) = log2 ^ 1 = 0
Similarly: because f (x) is an even function, f (- 2011) = f (2011) = f (2007) = f (2003) =. = f (3) = - f (1) = - log2 ^ 2 = - 1
So the final result is - 1

p. Whether q is a necessary and sufficient condition: P: F (- x) / F (x) = 1; Q: y = f (x) is even function

no
p：f（－x)/f(x)=1
Only when f (x) cannot be equal to 0, there is f (- x) = f (x)
q: If y = f (x) is an even function, then there is f (- x) = f (x), which has this relationship for all the values of X in the function definition domain
Therefore, P and Q are not necessary and sufficient conditions

p: F (- x) / F (x) = 1 Q: y = f (x) is an even function and P is q?

P is established, q is established
On the contrary, not necessarily
Because P has the premise that f (x) is not equal to 0
Therefore, there are sufficient and unnecessary conditions

p: Y = f (x) is even function, Q: F (- x) / F (x) = 1. What condition is p for Q?

1: Y = f (x) is an even function, where f (x) may take 0, then q is meaningless, so p is a non sufficient condition of Q;
2: If f (- x) / F (x) = 1, then f (x) ≠ 0, f (- x) = f (x), f (x) is even function, so p is a necessary condition of Q
To sum up, P is a necessary but not sufficient condition for Q

Let [log π (π a)] squared

Because the function f (x) = sin (x + a) + cos (x-a) is even
So f (a) = f (- a)
So sin2a + 1 = cos2a
So 2sina * cosa + 1 = 1-2 (Sina) ^ 2
So Sina * (Sina + COSA) = 0
So Sina * radical 2 * sin (a + π / 4) = 0
So Sina = 0 or sin (a + π / 4) = 0
So a = k π or (k-1 / 4) π (k is an integer)
Because [log π (π a)] squared

If f (x) = sin (Wx + AFA) + cos (Wx + AFA) (W > 0, absolute value AFA)

F (x) = sin (Wx + AFA) + cos (Wx + AFA) = √ 2 [√ 2 / 2 sin (Wx + AFA) + √ 2 / 2 cos (Wx + AFA)] (extract √ 2, root 2) = √ 2 [cos 45 ° sin (Wx + AFA) + sin 45 ° cos (Wx + AFA)] = √ 2 sin (Wx + AFA + 45 °) = √ 2 cos (Wx + AFA - 45 °) = √ 2 cos (Wx + AFA - 45 °) (because

Let f (x) = a * B, where vector a = (2cosx, 1), B = (cosx, √ 3sin2x), Find the minimum positive period and monotone decreasing interval of F (x)

Vector multiplication with coordinates a (x1, Y1) B (X2, Y2) a * b = X1 * Y1 + x2 * Y2, so here f (x) = a * b = 2cosx * cosx + 1 * √ 3sin2x = 2cosx ^ 2 + √ 3sin2x = cos2x + √ 3sin2x - 1 = 2 (1 / 2cos2x + √ 3 / 2sin2x) - 1 = 2 (sin π / 6 * Cos2

Let f (x) = AB, where vector a = (2cosx, 1), B = (cosx, root 3sin2x) (1) In the triangle ABC of the simple increasing interval (2) on [0, Wu], the opposite sides of angles a, B and C are a, B, C, and a ＋ B － C ＞ = AB, find the value of F (c)

F (x) = 2cos ^ x + Radix 3sin2x = cos2x + 1 + Radix 3sin2x = 2 (sin2xcospai / 6 + sinpai / 6cos2x) + 1 = 2Sin (2x + Pai / 6) + 1
Then the minimum positive period T = 2pai / 2 = Pai
-Pai/2+2kPai<=2x+Pai/6<=Pai/2+2kPai
-Pai/3+kPai<=x<=Pai/6+kPai
Therefore, the increasing interval on [0, Pai] is [0, Pai / 6] u [2pai / 3, Pai]
(2)a^2+b^2-c^2>=ab
cosC=(a^2+b^2-c^2)/2ab>=ab/2ab=1/2
Therefore, 0f (c) = 2Sin (2C + Pai / 6) + 1
Pai/6<=2C+Pai/6<=5Pai/6
Therefore, 1 / 2 < = sin (2C + Pai / 6) < = 1
Namely: 2 * 1 / 2 + 1 < = f (c) < = 2 * 1 + 1
That is: 2 < = f (c) < = 3

Let f (x) = m vector × n vector, where vector M = (2cosx, 1), n vector = (cosx, Geng, 3sin2x), X belongs to R. (1) find the minimum positive of F (x) Let f (x) = m vector × n vector, where vector M = (2cosx, 1), n vector = (cosx, Geng 3sin2x), X belongs to R. (1) Finding the minimum positive period and monotone decreasing interval of F (x)

f(x)=2*(cosx)^2-√3*sin2x
=1+cos2x-√3*sin2x
=2*[(1/2)*cos2x-(√3/2)*sin2x]+1
=2*cos(2x+π/3)+1
Minimum positive period: T = 2 π / 2 = π
Monotone decreasing interval:
kπ≤2x+π/3≤kπ+π/2
(3k-1)π/6≤x≤(6k-1)π/12 k∈z

Given the vector M = (Radix 3sin2x + 2, cosx), vector n = (1,2cosx), Let f (x) = vector m * vector n F = 4, B = 1. The area of ABC is the root of 2. Find the value of A First ask for F, first ask me to find out

F = radical 3sin2x + 2 + 2cosx ^ 2
=Radical 3sin2x + cos2x + 3
=2[cos(2x-60)] +3
f(a)=4
cos(2a-60)=1/2
2a-60=60
a=60
S = bcsina / 2 = radical 3 / 2
C=2
Using the cosine theorem:
a^2=b^2+c^2-2bccosa
=1+4-2=3
A = radical 3